2026-03-03 20:16:05 +0300 MSK

class Solution:
    def findKthBit(self, n: int, k: int) -> str:
        # Find the position of the rightmost set bit in k
        # This helps determine which "section" of the string we're in
        position_in_section = k & -k

        # Determine if k is in the inverted part of the string
        # This checks if the bit to the left of the rightmost set bit is 1
        is_in_inverted_part = ((k // position_in_section) >> 1 & 1) == 1

        # Determine if the original bit (before any inversion) would be 1
        # This is true if k is even (i.e., its least significant bit is 0)
        original_bit_is_one = (k & 1) == 0

        if is_in_inverted_part:
            # If we're in the inverted part, we need to flip the bit
            return "0" if original_bit_is_one else "1"
        else:
            # If we're not in the inverted part, return the original bit
            return "1" if original_bit_is_one else "0"